Two point electric charges of value `q` and `2q` are kept at a distance `d` apart from each other in air. A third charge `Q` is to be kept along the same line in such a way that the net force action on `q` and `2q` is zero. Calculate the position of charge `Q` in terms of `q` and `d`.

Explanation

To solve the problem, we need to find the position of charge \( Q \) such that the net force acting on charges \( q \) and \( 2q \) is zero. Let's denote the position of charge \( Q \) as \( x \) from charge \( q \) and the distance between \( q \) and \( 2q \) is \( d \).

### Step-by-Step Solution:

1. **Identify the Forces Acting on Charge \( Q \)**:
- The force acting on charge \( Q \) due to charge \( q \) is given by Coulomb's Law:
\[ F_{qQ} = k \frac{qQ}{x^2} \]
- The force acting on charge \( Q \) due to charge \( 2q \) is:
\[ F_{2qQ} = k \frac{2qQ}{(d - x)^2} \]

2. **Set Up the Force Balance Equation**:
- For the net force on charge \( Q \) to be zero, the forces must be equal in magnitude:
\[ F_{qQ} = F_{2qQ} \]
- Thus, we have:
\[ k \frac{qQ}{x^2} = k \frac{2qQ}{(d - x)^2} \]
- We can cancel \( k \) and \( Q \) (assuming \( Q \neq 0 \)):
\[ \frac{q}{x^2} = \frac{2q}{(d - x)^2} \]

3. **Simplify the Equation**:
- Cancel \( q \) (assuming \( q \neq 0 \)):
\[ \frac{1}{x^2} = \frac{2}{(d - x)^2} \]
- Cross-multiplying gives:
\[ (d - x)^2 = 2x^2 \]

4. **Expand and Rearrange the Equation**:
- Expanding the left side:
\[ d^2 - 2dx + x^2 = 2x^2 \]
- Rearranging gives:
\[ d^2 - 2dx - x^2 = 0 \]

5. **Solve the Quadratic Equation**:
- Rearranging gives:
\[ x^2 + 2dx - d^2 = 0 \]
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 2d \), and \( c = -d^2 \):
\[ x = \frac{-2d \pm \sqrt{(2d)^2 - 4 \cdot 1 \cdot (-d^2)}}{2 \cdot 1} \]
\[ x = \frac{-2d \pm \sqrt{4d^2 + 4d^2}}{2} \]
\[ x = \frac{-2d \pm \sqrt{8d^2}}{2} \]
\[ x = \frac{-2d \pm 2\sqrt{2}d}{2} \]
\[ x = -d \pm \sqrt{2}d \]

6. **Choose the Valid Solution**:
- Since \( x \) must be a positive distance, we take:
\[ x = (-1 + \sqrt{2})d \]

### Final Result:
The position of charge \( Q \) in terms of \( q \) and \( d \) is:
\[ x = (-1 + \sqrt{2})d \]